Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
Answer:
The heat transfer q = 18.32W
Explanation:
In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K
Please check attachment for complete solution and step by step explanation
Answer: It means "Do it yourself".
Explanation: You're welcome!
Answer with Explanation:
Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

Solving for 'U' we get

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%
The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2
i)
= 71% consolidation
ii)
= 45% consolidation
iii)
= 30% consolidation
Part b)
The degree of consolidation is given by

Thus a settlement of 50.46 centimeters has occurred
For time factor 0.7, U is given by

thus consolidation of 85.59 % has occured if time factor is 0.7
The degree of consolidation is given by
